Object A has mass mA = 8 kg and initial momentum pA,i = < 15, -8, 0 > kg · m/s, just before it strikes object B, which has mass mB = 11 kg. Just before the collision object B has initial momentum pB,i = < 2, 7, 0 > kg · m/s.
A) Consider a system consisting of both objects A and B. What is the total initial momentum of this system, just before the collision?
B) The forces that A and B exert on each other are very large but last for a very short time. If we choose a time interval from just before to just after the collision, what is the approximate value of the impulse applied to the two-object system due to forces exerted on the system by objects outside the system?
C) Therefore, what does the Momentum Principle predict that the total final momentum of the system will be, just after the collision?
D) Just after the collision, object A is observed to have momentum pA,f = < 13, 4, 0 > kg · m/s. What is the momentum of object B just after the collision?

Answer:

1.1mg/min

Explanation:

We are given that

Volume of solution=250 ml

Mass of amiodarone=1 g

Infusion rate=17 ml/hr

We know that

1 g= 1000 mg

Ratio of 1000g:250 ml=[tex]\frac{1000}{25}=4 mg/ml[/tex]

The concentration of solution=4mg/ml

Amiodarone infusing (mg/min)=[tex]\frac{infusion \;rate(ml/hr)\times concentration}{60}[/tex]

Because 1 hr= 60 minute

Amiodarone infusing=1.1mg/m

Hence, 1.1 mg/min of amiodarone is infusing.

The amount of amiodarone infusing per minute is 1.1 mg when calculated from the prescribed dosage of 1 gram in a 250 ml IV solution infused at 17 ml/hour.

The question revolves around the calculation of the amiodarone dosage being infused per minute through an intravenous (IV) therapy. Given the infusion rate of 17 ml/hour and a total amount of 1 gram (1000 mg) of amiodarone in 250 ml of Normal Saline, we can calculate the dosage per minute. First, find the concentration of amiodarone in mg/ml by dividing 1000 mg by 250 ml to get 4 mg/ml. Then, we calculate the amount infused per hour by multiplying this concentration (4 mg/ml) by the infusion rate (17 ml/hour). Finally, divide by 60 minutes to get the dosage per minute.

The calculation steps are as follows:

1. Amiodarone concentration = 1000 mg ÷ 250 ml = 4 mg/ml

2. Amiodarone per hour = 4 mg/ml × 17 ml/hour = 68 mg/hour

3. Amiodarone per minute = 68 mg/hour ÷ 60 minutes/hour = 1.13 mg/minute

The amount of amiodarone infused per minute is therefore approximately 1.13 mg, which rounds to 1.1 mg/minute when rounding to the nearest tenth.

Answer:21.3 m/s

Explanation:

Given

speed [tex]u=18.6 m/s[/tex]

mass of fish [tex]m_f=2.30 kg[/tex]

Altitude [tex]h=5.50 m[/tex]

Time taken to cover h

[tex]h=ut+\frac{at^2}{2}[/tex]

[tex]5.5=\frac{9.8\times t^2}{2}[/tex]

[tex]t^2=1.122[/tex]

[tex]t=1.05 s[/tex]

Vertical velocity after [tex]t=1.05 s[/tex]

[tex]v_y=0+gt[/tex]

[tex]v_y=9.8\times 1.05=10.38 m/s[/tex]

Horizontal velocity will remain same [tex]u=18.6 m/s[/tex]

Net velocity [tex]v_{net}=\sqrt{u^2+v_y^2}[/tex]

[tex]v_{net}=\sqrt{18.6^2+10.38^2}[/tex]

[tex]v_{net}=\sqrt{453.76}=21.30 m/s[/tex]

Answer:

Lead, Ethyl alcohol and water.

Explanation:

Specific heat capacity of a substance can be define as the quantity of heat that is absorbed by a substance needed to change the temperature of a unit mass of one kilogram of the substance by one kelvin

The ranking of the final temperature of the substances from highest to lowest is;

Lead > Ethyl alcohol > Water

We are told that water has a very high specific heat capacity.

Specific heat capacity of ethyl alcohol is 2.46 J/g.K which is about 2/5 of water

Specific heat capacity of lead = (1/30) of water

Now, the specific heat capacity is the heat required to raise the temperature of a substance.

This means that the higher the specific heat capacity, the lower the final temperature.

Now, from the samples given, we can say that;

Thus, lead will have the highest final temperature followed by ethyl alcohol, then water.

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Answer:

If the distance is doubled, the force of gravity between the two bodies is one-fourth as strong as before

Explanation:

The force of gravity between two bodies depends on the mass and distance. But we will focus on distance since that's what the question asks

Therefore, the force of gravity decreases as distance between the bodies increases.

Answer:

Distance in mm will be 0.3718 mm

Explanation:

We have given charge surface charge density [tex]\rho _s=5\mu c/m^2=5\times 10^{-6}\mu c/m^2[/tex]

We know that electric field due to surface charge density is given by

[tex]E=\frac{\rho _S}{2\epsilon _0}=\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}=2.824\times 10^5Volt/m[/tex]

We have given potential difference V = 105 volt

We know that potential difference is given by [tex]V=Ed[/tex]

So [tex]105=2.824\times 10^5\times d[/tex]

[tex]d=37.181\times 10^{-5}m=0.3718mm[/tex]

The correct answers are as follows:

1) The magnitude of the net force on the block at the moment it is released is given by Hooke's Law for springs in parallel, which states that the net force is the sum of the forces exerted by each spring. Since the block is pulled to the left, the force exerted by the left spring is to the right, and the force exerted by the right spring is to the left. Thus, the net force[tex]\( F \)[/tex] is:

[tex]\[ F = k_{\text{left}} \cdot x + k_{\text{right}} \cdot x \] \[ F = (31 \, \text{N/m} + 53 \, \text{N/m}) \cdot 0.27 \, \text{m} \] \[ F = 84 \, \text{N/m} \cdot 0.27 \, \text{m} \] \[ F = 22.68 \, \text{N} \][/tex]2) The effective spring constant [tex]\( k_{\text{eff}} \)[/tex] of the two springs in parallel is the sum of the individual spring constants:

[tex]\[ k_{\text{eff}} = k_{\text{left}} + k_{\text{right}} \] \[ k_{\text{eff}} = 31 \, \text{N/m} + 53 \, \text{N/m} \] \[ k_{\text{eff}} = 84 \, \text{N/m} \][/tex]

3) The period of oscillation [tex]\( T \)[/tex] for a mass-spring system is given by:

[tex]\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} \] \[ T = 2\pi \sqrt{\frac{7.4 \, \text{kg}}{84 \, \text{N/m}}} \] \[ T = 2\pi \sqrt{\frac{7.4}{84}} \] \[ T = 2\pi \sqrt{0.0881} \] \[ T = 2\pi \cdot 0.2968 \] \[ T \approx 1.86 \, \text{s} \][/tex]

4) The time it takes for the block to return to equilibrium for the first time is half of the period of oscillation:

[tex]\[ t = \frac{T}{2} \] \[ t = \frac{1.86 \, \text{s}}{2} \] \[ t \approx 0.93 \, \text{s} \][/tex]

5) The speed of the block as it passes through the equilibrium position can be found using the conservation of energy. The total mechanical energy is constant, so the potential energy at the release point is converted into kinetic energy at the equilibrium position:

[tex]\[ \frac{1}{2} k_{\text{eff}} x^2 = \frac{1}{2} m v^2 \] \[ k_{\text{eff}} x^2 = m v^2 \] \[ v^2 = \frac{k_{\text{eff}} x^2}{m} \] \[ v = \sqrt{\frac{k_{\text{eff}} x^2}{m}} \] \[ v = \sqrt{\frac{84 \, \text{N/m} \cdot (0.27 \, \text{m})^2}{7.4 \, \text{kg}}} \] \[ v = \sqrt{\frac{84 \cdot 0.0729}{7.4}} \] \[ v = \sqrt{\frac{6.1296}{7.4}} \] \[ v \approx \sqrt{0.8284} \] \[ v \approx 0.909 \, \text{m/s} \][/tex]

6) The magnitude of the acceleration[tex]\( a \)[/tex]of the block as it passes through equilibrium is given by Newton's second law:

[tex]\[ F = m \cdot a \] \[ a = \frac{F}{m} \] \[ a = \frac{22.68 \, \text{N}}{7.4 \, \text{kg}} \] \[ a \approx 3.065 \, \text{m/s}^2 \][/tex]

7) The position[tex]\( x(t) \) of the block at a time \( t \)[/tex] after it is released can be found using the equation for simple harmonic motion: [tex]\[ x(t) = A \cos(2\pi f t) \] \[ x(t) = 0.27 \cos\left(\frac{2\pi}{1.86} \cdot 1.06\right) \] \[ x(t) = 0.27 \cos(3.61\pi) \] \[ x(t) \approx 0.27 \cdot (-1) \] \[ x(t) \approx -0.27 \, \text{m} \][/tex]

8) The net force on the block at time [tex]\( t \)[/tex] is given by Hooke's Law, taking into account the position of the block:

[tex]\[ F(t) = k_{\text{eff}} \cdot x(t) \] \[ F(t) = 84 \, \text{N/m} \cdot (-0.27 \, \text{m}) \] \[ F(t) \approx -22.68 \, \text{N} \][/tex]

9) The total energy stored in the system[tex]\( E \)[/tex] is the potential energy at the maximum displacement, which is equal to the kinetic energy at the equilibrium position:

[tex]\[ E = \frac{1}{2} k_{\text{eff}} x^2 \] \[ E = \frac{1}{2} \cdot 84 \, \text{N/m} \cdot (0.27 \, \text{m})^2 \] \[ E = 42 \, \text{N/m} \cdot 0.0729 \, \text{m}^2 \] \[ E \approx 3.065 \, \text{J} \][/tex]

10) The period of oscillation is independent of the amplitude of the motion and depends only on the mass and the spring constant. Therefore, if the block had been given an initial push, the period of oscillation would not change. The correct answer is: the period would not change.

The amplitude is 2.00 cm. The initial phase angle, the maximum speed, and the wave equation can be found by substituting the given values into the relevant formulas.

The (a) amplitude of the wave is determined by the maximum displacement from the equilibrium position, which is 2.00 cm.

(b) The initial phase angle can be determined by the equation φ = -arccos(v₁/ωA), where v₁ is the initial speed of the element, ω is the angular frequency, and A is the amplitude. Here, ω=2π/T, T is the period. Therefore, φ = -arccos((-2.00 m/s)/(2π/(25.0 ms)*2.00 cm))

(c) The maximum transverse speed of an element of the string would be ωA, since the speed of the element is highest when it passes through its equilibrium position. Thus, the maximum speed = 2π/(25.0 ms)*2.00 cm.

(d) And lastly, the wave equation for the wave is Y(x,t) = Acos(kx + ωt + φ), where Y is the transverse displacement, A is the amplitude, k is the wave number, ω is the angular frequency, t is time, x is position, and φ is the phase constant. In this case, k = 2π/λ, where λ is the wavelength which can be found by dividing the speed of the wave by the frequency (v/f = λ).

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We determined the wave's amplitude to be approximately 3.33 cm, the initial phase angle to be π/3, and the maximum transverse speed to be about 8.37 m/s. The wave equation was found to be y(x, t) = 0.0333 sin(8.38x + 251.33t + π/3). Step-by-step explanations clarify each part.

Part a :

We start with the transverse wave function: y(x, t) = A sin(kx + ωt + φ). Given that at t = 0 and x = 0, the transverse position y is 2.00 cm, traveling downward with a speed of 2.00 m/s.

First, determine the frequency, and hence the angular frequency ω. The period T is given as 25.0 ms:

ω = 2π/T = 2π/(25.0 x 10⁻³) s ≈ 251.33 rad/s

The transverse speed of the string element is:

dy/dt = -Aω cos(φ) = -2.00 m/s (since it's traveling downward)

Given y(0, 0) = A sin(φ) = 0.020 m:

A = 0.020 m / sin(φ)

Using the transverse speed equation, we get:

-0.25133 A cos(φ) = -2.00 m/s

cos(φ) = 2.00 / (0.25133 A)

From both equations, sin²(φ) + cos²(φ) = 1, solving for A.

A = 2.00 cm x √(1 + (1/0.25133²)) ≈ 3.33 cm

Part b :

sin(φ) = 0.020 / 0.0333 ≈ 0.6

cos(φ) = 2.00 / (251.33 x 0.0333) ≈ 0.8

φ ≈ π/3

Part c :

The maximum transverse speed is given by Aω ≈ 3.33 cm x 251.33 rad/s ≈ 8.37 m/s.

Part d :

The general form of the sinusoidal wave traveling in the negative x-direction is:

y(x, t) = 0.0333 sin(kx + ωt + π/3)

Given that the wave speed v = 30.0 m/s:

k = ω/v = 251.33 rad/s / 30.0 m/s ≈ 8.38 m-1

Wave equation becomes:

y(x, t) = 0.0333 sin(8.38x + 251.33t + π/3)

Final answer:

The combined velocity of the two football players just after they collide and cling together is calculated to be 0.80 m/s in the direction of the first player's original motion, using the law of conservation of momentum.

Explanation:

To determine the velocity just after impact when two football players collide and cling together, we can use the law of conservation of momentum. The initial momentum of the system is the sum of the momentum of each player before they collide. For the first player with a mass of 95.0 kg and velocity of 6.00 m/s, and the second player with a mass of 115 kg and velocity of -3.50 m/s, the total initial momentum is:

(95.0 kg × 6.00 m/s) + (115 kg × -3.50 m/s) = 570 kg·m/s - 402.5 kg·m/s = 167.5 kg·m/s.

After the collision, the two players cling together and thus have a combined mass of 95.0 kg + 115 kg = 210 kg. The final velocity of the two players clinging together can be found by dividing the total initial momentum by the combined mass:

Final velocity = Total initial momentum / Combined mass = 167.5 kg·m/s / 210 kg = 0.80 m/s.

Therefore, the combined velocity of the two football players just after the collision is 0.80 m/s in the direction of the first player's initial motion.

Answer:

1.0326 m/s

Explanation:

mass of each ball, m = 155 g = 0.155 kg

mass of Dash and his skateboard, M = 65 kg

Speed of skateboard moving backward, v = 1.04 m/s

Let the speed of the balls in forward direction is u.

By using the conservation of momentum

Momentum in backward direction = Momentum in forward direction

(M + 3m) x u = M x v

(65 + 3 x 0.155) x u = 65 x 1.04

65.465 u = 67.6

u = 1.0326 m/s

Thus, the speed of balls in forward direction is 1.0326 m/s.

If Dash wants to move backward with a speed of 1.04 m/s, he should throw the balls forward with a velocity of about -0.996 m/s.

To calculate the velocity with which Dash should throw the balls to move backward, we can use the principle of conservation of momentum. According to this principle, the total momentum before throwing the balls should be equal to the total momentum after throwing the balls.

The initial momentum of Dash and his skateboard can be calculated as the product of their combined mass and velocity. The final momentum can be calculated as the sum of the momentum of Dash and his skateboard moving backward, and the momentum of the balls moving forward.

Using the formula for momentum (momentum = mass * velocity), we can set up the equation: (65 kg) * (-1.04 m/s) = (68 kg) * V, where V is the velocity with which Dash should throw the balls.

Solving for V, we find that Dash should throw the balls forward with a velocity of about -0.996 m/s to move backward with a speed of 1.04 m/s.

Answer:

a)v= 1.6573 m/s

Explanation:

a) Considering center of the disc as our reference point. The potential energy as well as the kinetic energy are both zero.

let initially the block is at a distance h from the reference point.So its potential energy is -mgh as its initial KE is zero.

let the block descends from h to h'

During this descend

PE of the block = -mgh'            {- sign indicates that the block  is descending}                

KE= 1/2 mv^2

rotation KE of the disc=  1/2Iω^2

Now applying the law of conservation of energy we have

[tex]-mgh = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2-mgh'[/tex]

[tex]mg(h'-h) = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex] ................i

Rotational inertia of the disc = [tex]\frac{1}{2}MR^2[/tex]

Angular speed ω =[tex]\frac{v}{R}[/tex]

by putting vales of  ω and I we get

so, [tex]\frac{1}{2}I\omega^2= \frac{1}{4}Mv^2[/tex]

Now, put this value of rotational KE in the equation i

[tex]mg(h'-h) = \frac{1}{4}(2m+M)v^2[/tex]

⇒[tex]v= \sqrt{\frac{4mg(h'-h)}{2m+M} }[/tex]

Given that (h'-h)= 0.5 m M= 360 g m= 70 g

[tex]v= \sqrt{\frac{4\times70\times 9.81\times 0.5}{140+360} }[/tex]

v= 1.6573 m/s\

b) The rotational Kinetic energy of the disc is independent of its radius hence on changing the radius there is no change in speed of the block.

Answer:

The speed of the block is 1.65 m/s.

Explanation:

Given that,

Radius = 12 cm

Mass  of pulley= 360 g

Mass of block = 70 g

Distance = 50 cm

(a). We need to calculate the speed

Using energy conservation

[tex]P.E=K.E[/tex]

[tex]P.E=mgh[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\times(\dfrac{v}{r})^2[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5Mr^2\times(\dfrac{v}{r})^2[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5M\times v^2[/tex]

[tex] K.E=\dfrac{1}{2}v^2(m+0.5M)[/tex]

Put the value into the formula

[tex]mgh=\dfrac{1}{2}v^2(m+0.5M)[/tex]

[tex]v^2=\dfrac{2mgh}{m+0.5M}[/tex]

[tex]v=\sqrt{\dfrac{2mgh}{m+0.5M}}[/tex]

[tex]v=\sqrt{\dfrac{2\times70\times10^{-3}\times9.8\times50\times10^{-2}}{70\times10^{-3}+0.5\times360\times10^{-3}}}[/tex]

[tex]v=1.65\ m/s[/tex]

(b), We need to calculate the  speed of the block

When r = 5.0 cm

Here, The speed of the block is independent of radius of pulley.

Hence, The speed of the block is 1.65 m/s.

Answer

given,

rotation of disk is equal to(θ) = 25.9

time taken = 5 s

moment of inertia of disk = 4 Kg.m²

disk come to rest in = 12 sec

a)  θ = 2 π x 25.9

    θ = 162.73 rad

    using equation of rotational motion

     θ =  ω₀t + 0.5 α t²

     162.79 =  0 + 0.5 x α x 5²

      α = 13.02 rad/s²

total torque acting

 [tex]\tau_{frictional} + \tau_{rotational} = I \alpha[/tex]

 [tex]\tau_{frictional} + \tau_{rotational} = 4\times 13.02[/tex]

 [tex]\tau_{frictional} + \tau_{rotational} =52.08[/tex]

again using the rotational equation

   [tex]\omega_f - \omega_i = \alpha\ t[/tex]

   [tex]\omega_f -0 = 13.02 \times 5[/tex]

   [tex]\omega_f = 65.1\ rad/s[/tex]

when only friction is acting angular acceleration

   [tex]\omega_f - \omega_i = \alpha\ t[/tex]

   [tex]0 -65.1 =\alpha\ 12[/tex]

   [tex]\alpha= -5.425\ rad/s^2[/tex]    

now torque due to friction

  [tex]\tau_{frictional} =- 4 \times 5.425[/tex]

  [tex]\tau_{frictional} =-21.7\ Nm[/tex]

now,

 [tex]-21.7+ \tau_{rotational} =52.08[/tex]

 [tex] \tau_{rotational} =73.78\ Nm[/tex]

The Coefficient of Performance (COP) of the heat pump is 2.22 and the rate at which it absorbs energy from the outdoor air is 1222 Watts.

The quality of a heat pump is judged by how much energy is transferred by heat into the warm space compared with how much input work is required. This measure is referred to as the Coefficient of Performance (COP). To calculate the COP of a heat pump which supplies energy at a rate of 8000 kJ/h for each kW of electrical power it draws, we have to convert all the units to the same base, which in this case will be watts (W).

1 kW = 1000 W and 8000 kJ/h = (8000*1000) J/3600 s = 2222 W

Hence, using the formula for the COP of the heat pump: COPhp = Qh/W, we substitute the given values and we get: COPhp = 2222W/1000W = 2.22. This means that for every 1 Watt of electricity the heat pump uses, it generates 2.22 Watts of heat for the house.

Additionally, the rate of energy absorption from the outdoor air is the difference between the rate of heat supply to the house and the electric power drawn, which is 2222W - 1000W = 1222W.

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The COP of the heat pump is 2.22, and the rate of energy absorption from the outdoor air is 1.22 kJ/s.

Determining the COP of a Heat Pump

The Coefficient of Performance (COP) of a heat pump is defined as the ratio of heat energy delivered to the heated space (Qh) to the energy input (W). In this case, we are given that the heat pump supplies 8000 kJ/h for each kW of electric power it draws.

Firstly, convert the supplied energy and power input to consistent units. Since 1 kW = 1 kJ/s, we have:

Energy supplied, Qh = 8000 kJ/h = 8000 / 3600 kJ/s = 2.22 kJ/s

Power input, W = 1 kW = 1 kJ/s

Apply the COP formula:

COP = Qh / W = 2.22 kJ/s / 1 kJ/s = 2.22

To determine the rate of energy absorption from the outdoor air (Qc), use the energy balance equation:

Qh = Qc + W

Solving for Qc:

Qc = Qh - W = 2.22 kJ/s - 1 kJ/s = 1.22 kJ/s

Thus, the COP of the heat pump is 2.22 and the rate of energy absorption from the outdoor air is 1.22 kJ/s.

Answer:

Earth's interior (Core)

Explanation:

The earth is comprised of 3 distinct layers namely the Core, the Mantle and the Crust, which are divided based on their composition as well as density.

The core of the earth is extremely very hot where the inner core remains solid and outer core acts a liquid. It is mainly comprised of iron, nickel and other siderophile elements.

A large amount of heat (energy) is radiated from this core region towards the surface of the earth. Due to this, the mantle rocks forms magma that creates the convection currents, where the hot and less dense magma rises upward and the cool and denser magma sinks to the bottom. This occurs continuously, as a result of which the lithospheric plates are forced to move over the less dense layer of asthenosphere.

Thus, the heat energy that drives the convection current in the mantle is provided from the interior (core) of the earth.

Answer:

earths core

Explanation:

Answer:

[tex]\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}[/tex]

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

[tex]m_{1} v_{1}=m_{2} v_{2}[/tex]

[tex]\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }[/tex]

[tex]\mathrm{v}_{1} \text { velocity before the collision }[/tex]

[tex]\mathrm{v}_{2} \text { velocity after the collision }[/tex]

[tex]\mathrm{m}_{1}=600 \mathrm{kg}[/tex]

[tex]\mathrm{m}_{2}=900 \mathrm{kg}[/tex]

[tex]\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}[/tex]

[tex]600 \times 5=900 \times v_{2}[/tex]

[tex]3000=900 \times v_{2}[/tex]

[tex]\mathrm{v}_{2}=\frac{3000}{900}[/tex]

[tex]\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}[/tex]

[tex]\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}[/tex]

Final answer:

After calculating with the conservation of momentum for an elastic collision, the 900 kg car will have a velocity of 3.81 m/s to the right after it is hit by the 600 kg car.

Explanation:

The question asks about the post-collision velocity of a 900 kg car that was initially stationary and was hit by a 600 kg car moving at 5 m/s which, after the collision, moved left at 0.714 m/s. Using the principle of conservation of momentum for elastic collisions, we can set up the equation as follows:

Initial momentum = Final momentum

(600 kg × 5 m/s) + (900 kg × 0 m/s) = (600 kg × -0.714 m/s) + (900 kg × v)

Solving for v (the velocity of the 900 kg car after the collision), we obtain:

3000 kg·m/s = -428.4 kg·m/s + 900 kg × v

v = (3000 kg·m/s + 428.4 kg·m/s) / 900 kg

v = 3.81 m/s

Thus, the velocity of the 900 kg car after the collision is 3.81 m/s to the right.

The correct answer is B. The base of the pyramid generally represents primary consumers.

Explanation

A biomass pyramid is a graphic representation of the biomass present in a unit area of various trophic levels, this graphic representation shows the relationship between biomass and the trophic level that quantifies the biomass available at each trophic level of an energy community at a specific time. In general, an ecosystem is represented in a pyramid in which the primary producers occupy the base of the pyramid because they have more biomass in a unitary area. An example of the organization of a biomass pyramid is an ecosystem in which caterpillars feed on oak trees; In turn, the caterpillars are consumed by a bluebird, which is consumed by a sparrowhawk. In this example, the oak tree is at the base of the biomass pyramid, because it feeds dozens of caterpillars, thanks to its massive biomass, and the sparrowhawk occupies the highest level of the pyramid. Therefore, it is incorrect to affirm that the base of the pyramid generally represents primary consumers, because the primary producers are generally at the base of the pyramid. So, the correct answer is B. The base of the pyramid generally represents primary consumers.

Answer:22,833.12 N

Explanation:

Given

Velocity of Jet [tex]v=201 m/s[/tex]

Velocity of Exhaust gases [tex]u=669 m/s[/tex]

Rate of intake of air [tex]\frac{\mathrm{d} M_a}{\mathrm{d} t}=44.1 kg/s[/tex]

Rate at which Fuel is burned is [tex]\frac{\mathrm{d} M_f}{\mathrm{d} t}=3.28 kg/s[/tex]

Rate of change of mass in Rocket [tex]\frac{\mathrm{d} M}{\mathrm{d} t}=\frac{\mathrm{d} M_a}{\mathrm{d} t}+\frac{\mathrm{d} M_f}{\mathrm{d} t}=44.1+3.28=47.38 kg/s[/tex]

Thrust on Rocket is given by

[tex]F=\frac{\mathrm{d} M}{\mathrm{d} t}u-\frac{\mathrm{d} M_a}{\mathrm{d} t}v[/tex]

[tex]F=47.38\times 669-44.1\times 201[/tex]

[tex]F=31,697.22-8864.1=22,833.12 N[/tex]

Answer:

3.16 i1

Explanation:

T1 = 300 k

I1 = i1

T2 = 400 k

i2 = ?

The intensity is directly proportional to the four powers of the absolute temperature.

I ∝ T^4

So, [tex]\frac{i_{2}}{i_{1}}=\frac{T_{2}^{4}}{T_{1}^{4}}[/tex]

[tex]\frac{i_{2}}{i_{1}}=\frac{400^{4}}{300^{4}}[/tex]

i2 = 3.16 i1

Thus, the intensity becomes 3.16 i1.

Answer:

C.O.P = 1.49

W = 335.57 joules

Explanation:

C.O.P = coefficient of performance = (benefit/cost) = Qc/W ...equ 1 where C.O.P is coefficient of performance, Qc is heat from cold reservoir, w is work done on refrigerator.

Qh = Qc + W...equ 2

W = Qh - Qc ...equ 3 where What is heat entering hot reservoir.

Substituting for W in equ 1

Qh/(Qh - Qc) = 1/((Qh /Qc) -1) ..equ 4

Since the second law states that entropy dumped into hot reservoir must be already as much as entropy absorbed from cold reservoir which gives us

(Qh/Th)>= (Qc/Tc)..equ 5

Cross multiple equ 5 to get

(Qh/Qc) = (Th/Tc)...equ 6

Sub equ 6 into equation 4

C.O.P = 1/((Th/Tc) -1)...equ7

Where Th is temp of hot reservoir = 493k and Tc is temp of cold reservoir = 295k

C.O.P = 1/((493/295) - 1)

C.O.P = 1.49

To solve for W= work done on every cycle

We substitute C.O.P into equ 1

Where Qc = 500 joules

1.49 = 500/W

W = 500/1.49

W = 335.57 joules

Answer:

0.674 s = t

Explanation:

Assuming that the door is completely open, exena need to rotate the door 90°.

Now, using the next equation:

T = I∝

Where T is the torque, I is the moment of inertia and ∝ is the angular aceleration.

Also, the torque could be calculated by:

T = Fd

where F is the force and d is the lever arm.

so:

T = 220N*1.25m

T = 275 N*m

Addittionaly, the moment of inertia of the door is calculated as:

I = [tex]\frac{1}{3}Ma^2[/tex]

where M is the mass of the door and a is the wide.

I  =[tex]\frac{1}{3}(750/9.8)(1.25)^2[/tex]

I = 39.85 kg*m^2

Replacing in the first equation and solving for ∝, we get::

T = I∝

275 = 39.85∝

∝ = 6.9 rad/s

Now, the next equation give as a relation between θ (the angle that exena need to rotate) ∝ (the angular aceleration) and t (the time):

θ = [tex]\frac{1}{2}[/tex]∝[tex]t^2[/tex]

Replacing the values of θ and ∝ and solving for t, we get:

[tex]\sqrt{\frac{2(\pi/2)}{6.9 rad/s}}[/tex] = t

0.674 s = t

Answer:

warmer

Explanation:

The law of conservation of energy tells us that energy cannot be created or destroyed, it can be transferred or converted from one from to another. In this question when the beer that is at room temperature is put in the fridge, it loses some heat energy. This heat energy is not destroyed, the fridge through  multiple processes eventually releases this heat to the room through pipes at the back which is why they are normally warm. the heat from the food inside is expelled to the room. It is not lost.

Answer:

thrust = 24795.87 N

Power  = 7.215 MW

Explanation:

given data

The speed of the jet v = 291 m/s

The air intake is dMa/dt = 63.8 kg/s

The fuel burn rate is dMf/dt = 1.21 kg/s

The speed of the exhaust gas is u = 667 m/s

to find out

thrust of the jet engine and the delivered power

solution

first we get here rate of mass change in the rocket that is  

rate of mass change dM/dt = 63.8  + 1.21

rate of mass change dM/dt = 65.01 kg/s

and

The thrust on the rocket will be here

thrust T = (dM/dt) u - (dMa/dt) v       ..................1

thrust T = (65.01) 667 - (63.8) 291

thrust T = 24795.87 N

and

now the power of the rocket will be here

Power  = speed × thrust        .................2

Power  = 291  × 24795.87

Power  = 7.215 MW

:Answer:

a. the initial angular velocity = 166.5rad/s

b. t= 135.7seconds

Explanation:

a. v = rw

angular velocity for 2870 rev, w = (2πN)/60 =2*3.142*2870rev)/60 =300.5rad/sec

w₁-w₂ = 300.5 - 133= 166.5rad/s

the initial angular velocity = 166.5rad/s

acceleration = change in velocity / time

b. 1 revolution = 2π

1 rad = 0.159rev

? = 2870 rev

=2870*1)/0.159 = 18050.3rad

the final angular speed = rev/time = rad/time

133 rad/s = 18050.3/t

133t = 18050.3

t= 135.7seconds

Answer:

It regulates the movement of proteins and RNAs into and out of the nucleus

Explanation:

The nuclear pore complex are protein channels connecting the outer membrane of the nucleus to the inner membrane of the nucleus. They securely regulates the almost all of the transport of protein and RNAs into and out of the nucleus.

The nuclear pore complex in eukaryotes is a protein complex that regulates the transportation of molecules between the nucleus and cytoplasm. It selectively allows passage of specific molecules, including ions, proteins, and RNA. It is integral to cellular functioning and to maintaining the cell’s genetic stability.

The nuclear pore complex found in eukaryotes regulates the traffic of molecules between the nucleus and cytoplasm. This rosette-shaped complex, composed of multiple proteins, allows for the selective passage of molecules such as ions, RNA, and proteins. Hence, it is crucial for the cell's functioning. For instance, RNA, which is created and spliced within the nucleus, needs to be transported to the cytoplasm for translation, and this transportation occurs through the nuclear pore complex. Moreover, the nuclear pore complex helps maintain the structure of the nucleus by allowing the passage of ions and molecules, ensuring the proper functioning of the nucleoplasm.

The composition of the nuclear pore complex makes it an efficient system for the transfer and regulation of certain molecules. Besides, it secures the cell's nucleus against unwanted, larger, or harmful substances that could potentially penetrate and damage the nucleus. Consequently, the nuclear pore complex contributes significantly to maintaining the cell's health and its genetic stability.

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Answer:East

Explanation:

Given

The truck is accelerating towards east along with crate and crate is not sliding.

Friction Force on the crate will act towards the east as friction Force always opposes the motion of an object. Also in this case, if friction force is absent then crate would have moved backward.

Thus static Friction will help the crate to move with truck.

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

[tex]\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}[/tex]

[tex]\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}[/tex]

[tex]\frac{3-y}{y}=\sqrt{\frac{7}{2}}[/tex]

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

Answer:

Answered

Explanation:

a) Two balls are at a distance of L/2 from the axis of rotation and one block at the center. ( center of rod).

therefore,

[tex]I=2\times m\frac{L}{2}^2[/tex]

[tex]I= \frac{1}{2}mL^2[/tex]

b) two balls at a distance L/4 at the from the axis and 1 ball at a distance 3L/4 from the from the axis.

[tex]I= 2\times m\times(L/4)^2 + m(\frac{3L}{4})^2[/tex]

= [tex]\frac{1}{16} mL^2(2+9)= \frac{11}{16}mL^2[/tex]

The moment of inertia of the system about an axis perpendicular to the rod and passing through the center of the rod is 1/2mL².

It should be noted that the moment of inertia of the system about an axis perpendicular to the rod and passing through the center of the rod will be calculated thus:

I = 1/2 × mL²/2

I = 1/2mL²

Also, the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one-fourth of the length from one end will be:

I = 2 × m × (L/4)² + m(3L/4)²

I = (2 + 9)1/16mL²

I = 11/16mL²

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Answer:

Is constant.

Explanation:

Isolated systems cannot exchange energy or matter outside the borders of the system.

An isolated system is a system that has no exchange of matter or energy with its environment. Despite changes in internal energy due to heat transfer or work done within the system, the total energy of an isolated system remains constant. The total energy of the system encompasses its kinetic energy, potential energy, and internal energy.

If a system is isolated, the total energy of the system is constant. This concept is part of the conservation of energy principle. An isolated system is one in which there is no exchange of matter or energy with the external environment. For instance, an ideal gas inside a thermally insulating and immoveable container would be an isolated system.

Though the internal energy of the system could change through processes such as heat transfer or work done on the system, the total energy remains unchanged. In other words, while energy itself might change forms within the system, the sum of kinetic, potential, and internal energy doesn't alter.

Consider a situation where heat transfer puts 159.00 J into the system, which raises the internal energy by 9.00 J. Despite how the energy was acquired, the system’s final state relates to its internal energy. Therefore, the initial and final total energy of the system remains same.

It's important to note that while total energy remains constant in isolated systems, in open systems (which can exchange energy and/or matter with its surroundings), the total energy can increase or decrease based on the work in/out of the system.

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Answer:

C. both forces have the same magnitude

Explanation:

Here the action force is equal to the reaction force in accordance with the Newton's third law of motion.

Also when we apply the conservation of momentum so that the momentum bullet and the momentum of the gun are equal and according to the second law of motion by Newton, we have force equal to the rate of change in momentum.

We have the equation for momentum as:

[tex]p=m.v[/tex]

Newton's second law is Mathematically given as:

[tex]F=\frac{dp}{dt}[/tex]

Momentum is constant and the reaction time is equal, so the force exerted will also be equal.

The correct answer is C. both forces have the same magnitude.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. This means that when the rifle fires the bullet, the force exerted by the rifle on the bullet is equal in magnitude to the force exerted by the bullet on the rifle.

 Let's denote the force exerted by the rifle on the bullet as [tex]\( F_{rb} \)[/tex]and the force exerted by the bullet on the rifle as[tex]\( F_{br} \).[/tex] According to Newton's third law:

[tex]\[ F_{rb} = -F_{br} \][/tex]

 The magnitudes of these forces are equal, even though they are in opposite directions.

 To calculate the magnitude of these forces, we can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it.

 Before the rifle is fired, both the bullet and the rifle are at rest, so their initial momenta are zero. After the rifle is fired, the bullet has a velocity of 300 m/s, and we can assume the rifle has a much smaller recoil velocity due to its much larger mass.

 The change in momentum of the bullet is:

[tex]\[ \Delta p_{bullet} = m_{bullet} \times v_{bullet} \][/tex]

[tex]\[ \Delta p_{bullet} = 0.00500 \, \text{kg} \times 300 \, \text{m/s} \][/tex]

[tex]\[ \Delta p_{bullet} = 1.5 \, \text{kg} \cdot \text{m/s} \][/tex]

 The change in momentum of the rifle is equal in magnitude and opposite in direction to the change in momentum of the bullet, assuming no other forces are acting on the system (like air resistance or friction):

[tex]\[ \Delta p_{rifle} = -\Delta p_{bullet} \][/tex]

[tex]\[ \Delta p_{rifle} = -1.5 \, \text{kg} \cdot \text{m/s} \][/tex]

 Since the time interval [tex]\( \Delta t \)[/tex] over which the forces are applied is the same for both the bullet and the rifle, the forces can be calculated using the impulse-momentum theorem:

[tex]\[ F_{rb} = \frac{\Delta p_{bullet}}{\Delta t} \][/tex]

[tex]\[ F_{br} = \frac{\Delta p_{rifle}}{\Delta t} \][/tex]

 Since [tex]\( \Delta p_{bullet} = -\Delta p_{rifle} \)[/tex], it follows that:

[tex]\[ F_{rb} = -F_{br} \][/tex]

 Therefore, the forces are equal in magnitude and opposite in direction, which is consistent with Newton's third law. The correct choice is C, both forces have the same magnitude.

Answer:

N = 516.56 N

Explanation:

By the means of a sum of forces on the x-axis:

[tex]N_b-N-f=0[/tex]  

Where [tex]N_b[/tex] is the force on her back and [tex]N_f[/tex] is the force on her feet:

[tex]N_b=N-f = N[/tex]  

On the y-axis:

[tex]Ff_b+Ff_f-m*g=0[/tex]

[tex]\mu_b*N_b+\mu_f*N_f-m*g=0[/tex]

[tex]\mu_b*N+\mu_f*N=m*g[/tex]

[tex](\mu_b+\mu_f)*N=m*g[/tex]

[tex]N=\frac{m*g}{\mu_b+\mu_f}[/tex]  using [tex]g=10m/s^2[/tex]

N = 516.56N

Answer:

Long wavelength

Explanation:

Wavelengths that corresponds to the bands of blue and red are strongly absorbed whereas the wavelengths that lie in the mid-range corresponds to green light that are absorbed weakly.

Fluorescence produced is always directed towards longer wavelengths of the spectra as compared to the corresponding spectra for absorption.